Please solve i’ll mark brainliest only do 6
1 answer:
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A) 
is continuous, so 
is also continuous. This means if we were to integrate 
, the same constant of integration would apply across its entire domain. Over 
, we have 
. This means that
For to be continuous, we need the limit as 
to match 
. This means we must have
Now, over
, we have 
, so 
, which means 
.
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
c) is increasing when 
, and concave upward when 
, i.e. when is also increasing.
We have
over the intervals 
and 
. We can additionally see that 
only on and .
d) Inflection points occur when
, and at such a point, to either side the sign of the second derivative 
changes. We see this happening at 
, for which 
, and to the left of we have decreasing, then increasing along the other side.
We also have along the interval 
, but even if we were to allow an entire interval as a "site of inflection", we can see that to either side, so concavity would not change.
For
Now, over
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
c)
We have
d) Inflection points occur when
We also have
Answer:
Tthe solution is in the attached file
Step-by-step explanation:
The step by step calculation is as shown in the attachment
